Significance The scale reading in part (a) of the figure above is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight: Fret = ma = 0 = F - w Fg= w= mg Fs = (75.0kg) (9.80 m/s2) = 735 N. Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In part (b), the scale reading is 735 N, which equals the person's weight. This is the case whenever the elevator has a constant velocity-moving up, moving down, or stationary. The solution to the previous example also applies to an elevator accelerating downward. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person's weight. If the elevator is in free fall and accelerating downward at g, then the scale reading is zero and the person appears to be weightless. Two Attached Blocks The figure below shows a block of mass 771 on a frictionless, horizontal surface. It is pulled by a light string that passes over a frictionless and massless pulley. The other end of the string is connected to a block of mass m2. Find the acceleration of the blocks and the tension in the string in terms of m1, m2, and g. 13 a m m1 - 11- y 5 a m2 x m2 W2 (a) (b) (a) Block 1 is connected by a light string to block 2. (b) The free-body diagrams of the blocks.